{"id":4870,"date":"2020-02-11T22:38:36","date_gmt":"2020-02-12T03:38:36","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/matg1013\/?p=4870"},"modified":"2025-12-13T09:55:54","modified_gmt":"2025-12-13T14:55:54","slug":"3eva2019tii_t4-completar-polinomio-de-interpolacion","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-3eva20\/3eva2019tii_t4-completar-polinomio-de-interpolacion\/","title":{"rendered":"3Eva2019TII_T4 completar polinomio de interpolaci\u00f3n"},"content":{"rendered":"\n<h2 class=\"wp-block-heading\">3ra Evaluaci\u00f3n II T\u00e9rmino 2019-2020. 11\/Febrero\/2020. MATG1013<\/h2>\n\n\n\n<p><strong>Tema 4<\/strong>. (25 puntos) Una funci\u00f3n f(x) en el intervalo [0,1] est\u00e1 definida por el trazador c\u00fabico natural S(x):<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> S_0(x) = 1 + 1.1186x + 0.6938 x^3 <\/span>\n\n\n\n<p class=\"has-text-align-center\">&nbsp;0.0 \u2264 x \u2264 0.4<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> S_1(x) = 1.4918 + 1.4516(x-0.4) + c(x-0.4)^2 +d(x-0.4)^3 <\/span>\n\n\n\n<p class=\"has-text-align-center\">0.4 \u2264 x&nbsp;\u2264 0.6<\/p>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> S_2(x) = 1.8221 + 1.8848(x-0.6) + <\/span>\n\n\n<span class=\"wp-katex-eq katex-display\" data-display=\"true\"> +1.3336(x-0.6)^2 - 1.1113(x-0.6)^3 <\/span>\n\n\n\n<p class=\"has-text-align-center\">0.6 \u2264 x \u2264 1.0<\/p>\n\n\n\n<p>Sin embargo, el papel donde se registraron los polinomios sufri\u00f3 un percance que no permite leer algunos valores para S<sub>1<\/sub>(x).<\/p>\n\n\n\n<p>a) Realice las operaciones necesarias para encontrar os valores: c, d<\/p>\n\n\n\n<p>b) Use el m\u00e9todo de Newton para resolver la ecuaci\u00f3n S(x) = 1.6<\/p>\n\n\n\n<p><strong>R\u00fabrica<\/strong>: plantear las condiciones(10 puntos), resolver el sistema (5 puntos), literal b (10 puntos)<\/p>\n","protected":false},"excerpt":{"rendered":"<p>3ra Evaluaci\u00f3n II T\u00e9rmino 2019-2020. 11\/Febrero\/2020. MATG1013 Tema 4. (25 puntos) Una funci\u00f3n f(x) en el intervalo [0,1] est\u00e1 definida por el trazador c\u00fabico natural S(x): &nbsp;0.0 \u2264 x \u2264 0.4 0.4 \u2264 x&nbsp;\u2264 0.6 0.6 \u2264 x \u2264 1.0 Sin embargo, el papel donde se registraron los polinomios sufri\u00f3 un percance que no permite [&hellip;]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"wp-custom-template-entrada-mn","format":"standard","meta":{"footnotes":""},"categories":[28],"tags":[60],"class_list":["post-4870","post","type-post","status-publish","format-standard","hentry","category-mn-3eva20","tag-interpolacion-polinomica"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4870","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=4870"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4870\/revisions"}],"predecessor-version":[{"id":17700,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/4870\/revisions\/17700"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=4870"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=4870"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=4870"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}